# Tag Info

10

Short answer - yes, everything in the circuit can contribute. But usually, an ohmmeter is zeroed with the probes in place - in other words, whatever resistance the probes represent is taken out by the meter. There are two other factors that play a role, especially when you try to measure small resistance. The first of these is contact resistance: it is ...

8

If you have an excess of electron in your body, your hair might stand on end and you might feel a bit negative (I couldn't help that pun), and you should probably avoid touching people or metal object if you don't want a static shock, but other than that, it's mostly harmless. The real danger comes from flowing electrons. Because the body basically runs on ...

6

Typically this is explained by the saying, "current kills." It's not the charge (or potential above ground) that a body attains that hurts biological systems, it's the current that flows through them and either 1) heats them or 2) disrupts important electrical signals in the body. Heating damage occurs and can "cook" (cause 1st, 2nd, or 3rd degree burns ...

3

Let me first take a little detour away from this circuit to particle accelerators. If you have some electrons in vacuum and a potential set up between two points (exactly the same as saying you have an electric field set up) you can accelerate your electrons. If you move a single electron through $1V$ of potential the electron gains $1eV$ of energy where ...

3

Imagine a free-standing battery (not connected to any wires) and take a closed loop through the battery, out one terminal, and back in the other terminal. The total work done in moving a test charge around that loop must vanish. For this to happen, the change in electric potential outside of the battery must equal the negative of the EMF change within the ...

3

should the voltage be the input voltage or the voltage across that specific resistor The latter. generally when you want to know something about a particular component, you work with the conditions applying at the boundary of that specific component. I have a desk lamp with a LED. The other end of the lamp plugs into a 240 V AC power outlet. If I want ...

2

The current is the conventional current in the opposite direction to the electrons current but they are the same thing , the current is the same in series connections. If the two lamps are identical they will give the same amount of light.

2

Since the lamps are in series, the electric current through each is identical; all of the current out of one lamp is in to the other lamp; if there is a flow through one, there is a flow through the other. It cannot be that there is a flow through one and not the other (in the context of this simple model). Thus, if the lamps are identical, their ...

2

My guess; you are mixing up quadripoles and quadrupoles. Quadripoles are two-port networks used in electric circuit analysis. The original German word is "Vierpol Theorie", which means Four-pol because of 4 Poles. https://en.wikipedia.org/wiki/Two-port_network Quadrupoles are related to multipole expansion used in electromagnetic, atomic orbital,.. theory. ...

2

Ok so I first have taken the diagram from the wikipedia page for reference and put it here. Now if you are happy with the idea of how the potential divider works... .... then I hope that you can see that $R_1$ and $R_2$ in the Wheatstone Bridge diagram form a potential divider and there is another potential divider with $R_3$ and $R_x$ - and the points ...

2

The whole electrical power grid is connected to ground. I don't know the details of other regions, but if you are in North America, the two current carrying conductors in a residential electrical outlet are called "hot" and "neutral". The "neutral" conductor is connected to the Earth at many places. If your bare feet touch wet Earth, and your hand touches ...

2

The thing to watch out for here is that it is easy to get into the habit of using sloppy language and/or notion for these things. I'll try to be precise here. It is correct to use the above relationship for the instantaneous power developed by a resistive component in a AC circuit $$P(t) = \frac{(V(t))^2}{R} \,.$$ Of course, that power will fluctuate ...

2

This is a common question. The issue is that the "Q" in $i = dQ/dt$ is not the same as the $Q$ that represents the charge on the capacitor. The variable $Q$ in use here is simply the charge on the capacitor. No problem. When the capacitor discharges the quantity of charge that is introduced into the circuit after a time $\delta t$ has elapsed is ...

2

I recently answered a similar question here. The ideal capacitor equation $$i_C = C\frac{dv_C}{dt}$$ assumes the passive sign convention which means that the reference direction for $i_C$ is into the positive labelled terminal. When you write $$iR = v_C$$ it is necessarily the case that $$i_C = - i$$ To see this, assume that both positive labelled ...

2

Take a capacitor and put it across a battery. There will be a transient current as the electrons go towards the anode . This happens very fast and the current is small. If you short the capacitor with a wire, the battery will empty all its charge on the short, which, depending on the battery can really be damaging. Your body accumulates some charge which ...

2

Your argument that the energy should radiate away would be true if your inductor were a good antenna, in which case it would be a bad inductor! The problem is an impedance mismatch: The inductor produces a magnetic field (which stores the energy you inquire about), but little electric field. That is the wrong ratio, or impedance, to couple to the vacuum ...

2

Kirchoff's voltage law. Quoting Wikipedia, The directed sum of the electrical potential differences (voltage) around any closed network is zero, or: More simply, the sum of the emfs in any closed loop is equivalent to the sum of the potential drops in that loop, or: The algebraic sum of the products of the resistances of the conductors ...

1

A few points of confirmation/correction: Yes the electrons flow in a direction opposite to the "conventional" current. No there is no "deficiency" if electrons - rather they have a different "potential" which is caused by the chemical reactions in the battery. No you don't have to invoke "surface charges" in the wire in order to understand current - ...

1

But isn't this solution wrong? I think it should be wrong, because we are talking about AC here; That is, 220V is peak voltage, not mean voltage. I think that mean voltage is something below 220V, thus resistance is going to be below 484 Ohms. If you do not live in central or north america, 220V will be RMS, so your teacher is right.

1

Trying to address this misconception: I start of with a resistance of 1 ohm by the wire and 6 amperes which result in 6 volts. When I meet the resistor however the resistance increases to let's say 3. Does the current decrease at the same rate the resistance increases? So if the resistance goes down to 3 will the current be 2 so that in the end I have a ...

1

So what exactly happens to the potential inside the resistor ? Unlike the ideal conductors, for which an electric field cannot exist inside, there is an electric field through the resistor body when there is a current through. And, as you may know, the rate of change in electric potential is related to the value of the electric field. Thus, the ...

1

My question is that if we repeat the above procedure in vacuum, would the sparks be produced? Not in perfect vacuum, because there is no gas to ionize. An understanding of this behavior can be attained by studying Paschen's Law (see http://en.wikipedia.org/wiki/Paschen%27s_law), which describes the breakdown voltage for given distance and pressure.

1

If you are not in a complete electrical circuit, any electric shock caused by touching a charged object or wire is brief. These "static shocks" are slightly painful, but they are (rarely) dangerous or fatal. I'm sure you've experienced a minor static shock. By wearing insulating footwear, you break a complete circuit and forbid a flow of electricity from ...

1

The bleeder resistor is across the capacitor so they have identical voltage across.

1

First let's establish the situation in which the result actually holds. Voltage itself is only well defined in electrostatics, and this result only holds in a steady state. In an ideal battery, there is no energy loss inside the battery during operation, and in the steady state just as much charge flows into the battery as flows out of the battery, and ...

1

First equation: power is defined as energy per unit time. Second equation: if a current flows through a circuit, the power dissipated is the product of voltage and current. This is because the voltage describes the energy each electron is given to traverse the circuit, and the current describes the number of electrons that travel the circuit per unit time. ...

1

First of all, I'll write the energy with the letter $U$ to not confuse with the electric field $E$. By definition of power, $P=\frac{dU}{dt}$. Now, where is this energy in the equation coming from? $$U=\int \vec{F}\cdot \vec{dl} = q \int \vec{E}\cdot \vec{dl} = qV$$ This is the amount of energy gained by the charge when moving across an electric potential ...

1

note: I accidentally thought OP was asking about a series $LC$, not a series $LCR$. Including the $R$ changes the results here by making the infinities turn into large finite values. Suppose you hook your series $LC$ circuit up to a voltage source with frequency dependent phasor $\tilde{V}_s(\omega)$. Intuition First let's guess what happens. At low ...

1

The resolution is that currents can change instantaneously. Remember that the reason that the current in a single solenoid cannot change instantaneously is that a change in current causes a change in flux, which causes an electromotive force (emf) to oppose the original change in current. However, suppose now there are two currents, as in your example. ...

1

There is no need for taking the mod of the charge. The voltage has the same sign as the charge. So if you start out with +20 µC on one capacitor (they give the + sign for a reason - so that's the side where we will put the positive charge) and +60 µC on the other (from $Q=CV$) then it follows that the total of redistributed charge is 80 µC as you correctly ...

Only top voted, non community-wiki answers of a minimum length are eligible